The photoelectric effect and Compton scattering show that electromagnetic waves sometimes exhibit particle-like properties. In 1923, de Broglie proposed that matter, which we normally think of as made up of particles like electrons, protons and neutrons, should have wave-like properties. In other words, everything has both wave- and particle-like properties.

Obviously, electromagnetic radiation does behave like a wave in many circumstances and matter behaves as if it were made up like particles in many circumstances, i.e., all circumstances where classical physics provides an adequate account. Therefore, whether a system exhibits wave-like or particle-like properties depends on the experiment that we are doing. This is known as wave-particle duality.

For a photon, we have $E = pc$ and the quantum postulate says that $E = h\nu$. Combining these gives $p = h\nu /c = h/\lambda$, or $\lambda = h/p$.

de Broglie proposed that the same relation should hold for matter particles, so a matter particle with momentum $p$ is associated with a wave of wavelength \[\boxed{\lambda = \frac{h}{p}.}\] This is known as the de Broglie wavelength of the particle.

de Broglie is a French name and can only be pronounced correctly by native French speakers. If you are a native English speaker then any way you choose to pronounce it is definitely wrong.

More generally, momentum $\vec{p}$ is a vector and the direction of propagation and wavelength of a wave is described by a wave vector $\vec{k}$ with magnitude given by the wave number $k = \frac{2\pi}{\lambda}$, so we want to define a relationship between the vector momentum and the wave vector. de Broglie's postulate tells us that the wave number is given by \[k = \frac{p}{\hbar},\] where \[\boxed{\hbar = \frac{h}{2\pi}.}\] The constant $\hbar$ is called the modified Planck's constant, but many people refer to both $h$ and $\hbar$ as Planck's constant, probably because $\hbar$ appears in the equations of quantum mechanics more frequently than $h$ and it would be tedious to keep saying “modified” the whole time. To avoid confusion, it is common to refer to $\hbar$ as $h$-bar. Although this is a boring name, it does lead to many extremely funny physics jokes such as “Where did the quantum physicist go for a drink after work?” and, well, I think you can guess the punchline.

Returning to more important matters, we need to upgrade the de Broglie wave number into a vector by choosing a direction. The simplest thing to do is to posit that the wave propagates in the same direction as the particle, so it has the same direction as the momentum. This gives \[\boxed{\vec{k} = \frac{\vec{p}}{\hbar}}.\] This makes sense if we imagine that a particle is somehow accompanied by a wave, because then the wave ought to be propagating in the same direction as the particle. The actual relationship between the wave and the particle, and whether either of them actually exist in the sense that we are used to from classical physics, is still controversial. We will discuss this in more detail later in the course.

Before computing the de Broglie wavelength of the electron, a word on units. Since Planck's constant and the masses of elementary particles are so small, the typical masses and energy scales involved in quantum mechanics are very small. Therefore, it is common to measure masses and energies in electron volts ($\text{eV}$) rather than the SI unit Joules ($\text{J}$).

The electron volt is a unit of energy, defined to be the change in electric potential energy of an electron as it moves across a potential difference of $1\,\text{V}$. Thus, the conversion factor is

\[1\,\text{eV} = 1.062\times 10^{-19}\,\text{J}\]

The mass of the electron is $9.109 \times 10^{-31}\,\text{kg}$. From special relativity, we know that mass and energy are equivalent. The rest energy of the electron is $m_e c^2$. In SI units this is $(9.109 \times 10^{-31}\,\text{kg})(2.998\times 10^{8}\,\text{ms}^{-1}) = 8.187\times 10^{-14}\,\text{J}$. Because these numbers are so small, it is common to covert the rest energy to electron volts, which gives $(8.187\times 10^{-14}\,\text{J})/(1.062\times 10^{-19}\,\text{JeV}^{-1}) = 5.11\times 10^{5}\,\text{eV} = 0.511\,\text{MeV}$. Somewhat confusingly, physicists will often say that the mass of the electron is $0.511\,\text{MeV}$, but it is important to remember that this really means the rest energy of the electron, so $m_e c^2 = 0.511\,\text{MeV}$.

When asked to do a calculation where masses and energies are given to you in electron volts, it is always possible to convert everything into SI units first and then do the calculation. However, it is also possible to do the calculation directly in electron volts using the fact that $hc = 1.240 \times 10^{-6} \,\text{eVm}$. The advantage of this is that the exponents in the calculation have smaller magnitude, so it is less likely that you will get a number that is too small/large for your calculator to handle during the calculation. To illustrate this, let's calculate the do Broglie wavelength of an electron with kinetic energy $54\,\text{eV}$ both ways: first by converting to SI units and then keeping everything in electron volts.

Since $54\,\text{eV} \ll 5.11\times 10^{5}\,\text{eV}$, the kinetic energy is a lot smaller than the rest energy of the electron, so using non-relativistic physics should be fine. The kinetic energy of the electron is \[K = \frac{1}{2}m_e v^2,\] where $v$ is the speed of the electron, and since the momentum is given by $p=m_e v$, we can write this as \[K=\frac{p^2}{2m_e}.\] This can be rearranged to \[p=\sqrt{2m_eK},\] and since the de Broglie wavelength is $\lambda = h/p$, we have \begin{equation} \label{debrogkinetic} \lambda = \frac{h}{\sqrt{2 m_e K}}. \end{equation}

First, let's do the calculation in SI units. We have to convert $54\,\text{eV}$ into Joules, which gives $(54\,\text{eV})(1.602\times 10^{-19}\,\text{JeV}^{-1}) = 8.651\times 10^{-18}\,\text{J}$. Substituting this into equation \eqref{debrogkinetic} then gives \[\lambda = \frac{6.626 \times 10^{-34}\,\text{Js}}{\sqrt{2(9.109 \times 10^{-31} \,\text{kg})(8.651\times 10^{-18}\,\text{J})}} = 1.67\times 10^{-10}\,\text{m} = 0.167\,\text{nm}.\]

If we instead, want to do the calculation in electron volts, then the first step is to multiply equation \eqref{debrogkinetic} by $c$ on the top and bottom to give \[ \lambda = \frac{hc}{\sqrt{2 m_e c^2 K}}.\] The reason for doing this is so that we have the rest energy of the electron $m_e c^2$ in the denominator rather than the mass. Doing this has also put $hc$ in the numerator, so we can use $hc = 1.240 \times 10^{-6} \,\text{eVm}$. Then, we have \[\lambda = \frac{1.240 \times 10^{-6}\,\text{eVm}}{\sqrt{2(0.511 \times 10^{6} \,\text{eV})(54\,\text{eV})}} = 1.67\times 10^{-10}\,\text{m} = 0.167\,\text{nm}.\]

We get the same answer, but the exponents involved in the calculation (${6}$ and ${-6}$) are much smaller in magnitude than the $-34$, $-31$ and $-18$ that we had when doing the calculation in SI units.

In this course, you may do calculations either in SI units or electron volts, you may report energies in Joules or electron volts, and masses in kilograms or electron volts (remembering that a mass given in electron volts is really a rest energy). The calculations in electron volts are often a bit simpler, and the trick to doing them is to multiply and divide equations by whatever factor of $c$ is needed to convert all the masses into rest energies. What you should never do however is mix the two methods. If you are using SI units then you need to convert every mass into kilograms, every energy into Joules, and always use $h$ in units of $\text{Js}$. If you are using electron volts then you need to have every mass (rest energy) in electron volts, every energy in electron volts, and always use $hc$ in units of $\text{eVm}$. If you mix the two then you will get an incorrect answer because your units will be nonsensical.

If matter particles sometimes behave like waves then it ought to be possible to detect wave interference effects like diffraction and double-slit interference. At this point, you should do the in-class activity to determine the typical de Broglie wavelengths of matter particles in order to understand why it is difficult to detect these interference effects.

In 1927, Davisson and Germer scattered a monoenergetic beam of electron with energy $54\,\text{eV}$ off a slab of Nickel crystal. They observed the distribution of the intensity of the scattered electrons as they varied the scattering angle. The following figure illustrates their setup.

They found an interference

1. For this activity you will need Planck's constant $h = 6.626 \times 10^{-34}\,\text{m}^2\text{kgs}^{-1}$.

   1. Carbon has atomic mass $\approx 12 \,\text{u}$ where $1\,\text{u}= 1.661\times 10^{-27}\,\text{kg}$. Estimate the de Broglie wavelength of a Carbon 60 molecule travelling at $166\,\text{ms}^{-1}$.
   2. Estimate the de Broglie wavelength of a person of mass $70\,\text{kg}$ running at $10\,\text{ms}^{-1}$.
   3. Why is it hard to observe interference of macroscopic objects?